Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}3x+y &= -4 \\ 6x-5y &= 8\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $6x = 5y+8$ Divide both sides by $6$ to isolate $x$ $x = {\dfrac{5}{6}y + \dfrac{4}{3}}$ Substitute this expression for $x$ in the first equation. $3({\dfrac{5}{6}y + \dfrac{4}{3}}) + y = -4$ $\dfrac{5}{2}y + 4 + y = -4$ Simplify by combining terms, then solve for $y$ $\dfrac{7}{2}y + 4 = -4$ $\dfrac{7}{2}y = -8$ $y = -\dfrac{16}{7}$ Substitute $-\dfrac{16}{7}$ for $y$ in the top equation. $3x- \dfrac{16}{7} = -4$ $3x-\dfrac{16}{7} = -4$ $3x = -\dfrac{12}{7}$ $x = -\dfrac{4}{7}$ The solution is $\enspace x = -\dfrac{4}{7}, \enspace y = -\dfrac{16}{7}$.